# Find the integer closest to $\ln(2013)$

I encounter such a problem, in a Maths contest, to find out the closest integer to $\ln(2013)$, without using a calculator. I really get stuck.

I tried to turn $\ln(2013)$ into $\ln(3)+\ln(11)+\ln(61)$, but nothing valuable obtained. I applied also Taylor series of natural log but it doesn't work. Any suggestions are welcomed.

• I thought this was a rather weird question as given, and seeing the different answers posted and how all of them use the damn symbol $\,\approx\,$ , I think it should/must be given some rational approximations to some values of the logarithm, like $\,\log 2\approx 0.7\,$ and etc. Without this assumption anyone can use almost any "approximation" and things get murky... Feb 28, 2013 at 11:58
• Thanks for your opinion. I know $\ln(2)$. But it's really unexpected that they use $\ln(10)$ and $e^3$. I ask the question in this forum now and naturally you find it weird. However it's a question in a contest, so I realise every single method will do, as long as they can remember the estimation. For me the simpler the estimation used is, the nicer the question is. Feb 28, 2013 at 14:14
• @DonAntonio I think the ln(2)≈0.7 approximation is adequate in this case, because we are looking for an integer, we need only one (or at most 2) significant digits for the computation. Feb 28, 2013 at 15:10
• This can clearly be solved simply by knowing the approximate value of $e$ and multiplying decimals (see Alex Jordan's answer). It is remarkable to me that this simple approach is not considered obvious; perhaps multiplying decimals is no longer something that is ever done "without using a calculator"! Mar 7, 2013 at 9:21

$2013$ is "very" close to $2048=2^{11}$. So how about $$2013=e^x=2^y$$ where $y$ is effectively equal to $11$. Then $x=y\ln 2$ and $\ln 2$ is famously equal to $0.7$. Then $$\ln(2013)\approx 11\cdot 0.7=7.7$$ giving an answer of $8$.

• Oh drats. I wrote a similar argument. Only that $\ln 2$ is not famously equal to any rational number, in particular not to $0.7$. Feb 28, 2013 at 6:08
• right, it's a joke... Feb 28, 2013 at 6:09
• Sort of like how $\pi^4 + \pi^5 = e^6$, etc, I guess: en.wikipedia.org/wiki/…
– Neal
Feb 28, 2013 at 6:15
• I've memorized $\ln 2$ to 3 decimal places. Feb 28, 2013 at 12:59
• This method seems a bit dangerous, given the 'correct' answer turns out to be less than 7.61. We wouldn't want to get too close to 7.5 May 8, 2013 at 12:35

Note that $2013$ is nearly $2048$ which is $2^{11}$.

Also note that $\ln(2013)=\log_2(2013)\cdot\ln 2$. Since $\log_2(2013)$ is nearly $\log_2(2048)=11$ and $\ln 2$ is roughly $0.693\approx 0.7$ we have that $\ln(2013)$ is roughly $11\cdot0.7\approx 7.7\approx 8$.

• Another useful rule of thumb is that to within 1%, $\ln x + \log_{10} x \approxeq \log_2 x$. So $\log_2 2013 \approxeq 11$ and $\log_{10} 2013 \approxeq \log_{10} 2000 = 3 + \log_{10} 2 \approxeq 3.3$. So $\ln 2013 \approxeq 7.7$. If you've forgotten an approximation for $\log_{10} 2$, you can use the fact that $\sqrt{\sqrt{10}} \approxeq 1.78$, so $\log_{10} 1.78 \approxeq 0.25$. Feb 28, 2013 at 6:17

Without remembering logs (while it may be useful to recall some),
Note $2 < e < 3$ and $2^{10} < 2013 < 3^7$
So if $e^x = 2013$, we must have $7 < x < 10$

Hence $e^\frac{x}{11} = 2013^\frac{1}{11} = (2048 - 35)^\frac{1}{11} = 2(1 - \frac{35}{2048})^\frac{1}{11}$

Now we have $\frac{x}{11} < 1$ and can approximate without fear of losing much accuracy using:

$1 + \dfrac{x}{11} + \dfrac{x^2}{242} \approx 2 - \dfrac{2\cdot 35}{11 \cdot 2048}$

$x^2 + 22 x \approx 241$
$(x+11)^2 \approx 362$
or $x \approx 8$

• Neat method, but quite complicated comparing to remembering that $\ln 2\approx 0.691$. Feb 28, 2013 at 7:18
• Agreed. Had a friend who memorised log and antilogs of many numbers, which was quite useful, till calculators became common. Still remembering a few is good, and so is Pseudonym's trick. Feb 28, 2013 at 7:28

$\ln 3$ is a little greater than $1$. In fact you can use $\ln (1+x)\approx x$ with $\frac 3e \approx 1.1$ to get $\ln 3 \approx 1+ \ln 1.1 \approx 1.1$

Maybe you know that $\ln 10 \approx 2.3$, so $\ln 11=\ln 10 + \ln 1.1 \approx 2.4$

Then $\ln 61 \approx \ln 2 + \ln 3 + \ln 10 \approx 0.7+1.1+2.3 =4.1$.

Summing it all up, we have $1.1+2.4+0.7+1.1+2.3=7.6$ and I would say $8$, though we were uncomfortably close to $7.5$. In fact $\ln 2013 \approx 7.607$ so the approximations were quite good.

Afterthought: even easier is $\ln 2000 \approx 0.7+3\cdot 2.3 = 7.6$ and the extra factor $1.006$ only adds $0.006$

One of my useful memorized rough approximations is $e^3\approx20$, and I know that $20$ is a slight underestimate. So $e^6$ is a bit over $400$, and $e^9$ is a bit over $8000$. That means that the choice is between $7$ and $8$. $400$ is too small by a factor of about $5$, and $8000$ is too big by a factor of only about $4$, so it’s $8$, though not by a whole lot. (And sure enough, it turns out to be about $7.61$.)

Clearly the answer is not all that large. So why not just multiply out powers of $e$ to a reasonable number of digits, like 3? Without too much work you'll hit close to $2013$ soon enough. All calculations below are by hand with either two or three digits preserved.

$$e\approx2.72$$ $$e^2\approx7.40$$ $$e^4=(e^2)^2\approx54.8$$ $$e^8=(e^4)^2\approx3000$$

Back pedal

$$e^6=(e^4)(e^2)\approx406$$ $$e^7=(e^6)(e)\approx1100$$

So with $e^7\approx1100$ and $e^8\approx3000$ we must judge where $2013$ falls.

$$1100\rightarrow(\times\approx1.8)\rightarrow2013\rightarrow(\times\approx1.5)\rightarrow3000$$ shows us that 2013 is relatively closer to 3000 than 1100. So we'd say the answer is 8. A formal proof would require more care paid to error bounds on all of the estimation ($\approx$).

• Another approach to the second half is to memorize the value of $e^{1/2}$ and use that to locate the right exponent. Feb 28, 2013 at 9:18
• That would work. It feels like that would be an even less likely thing to have memorized than some of the log values in other answers. I was shooting for an approach that used more common trivia. Feb 28, 2013 at 19:27
• Quite true, though you can always approximate it with a couple Newton-Raphson iterations, so you don't strictly have to memorize it. Mar 1, 2013 at 0:53

$2013$ is "very" close to $2000=2\cdot10^3$. So how about $$\ln(2013)\approx \ln(2000)$$ $$\ln(2000) = \frac{\log(2000)}{\log(e)}$$ remembering $$\log(e) = \frac{1}{\ln(10)}$$ then we have $$\ln(2000) = \log(2000)\cdot\ln(10)$$ $$= \log(2\cdot 10^3)\cdot\ln(10)$$ $$= (\log(2) + 3\cdot\log(10))\cdot\ln(10)$$ $$\approx (0.3 + 3)\cdot 2.3$$ $$= 3.3 \cdot 2.3$$ $$\approx 7.6$$ $$\approx 8$$

Well $\,\ln(2)\approx 0.69315\,$ and $\,\ln(10)\approx 2.302585\$ so that : $$\ln(2013)=\ln(2)+\ln(10^3)+\ln(1.0065)\approx 0.69315+3\cdot 2.302585+0.0065$$ (with an error of order $\frac 12 0.0065^2$)
getting : $$\ln(2013)\approx 7.6074$$ (of course $\ \ln(2013)\approx 0.7+3\cdot 2.3\approx 7.6\$ was enough here...)

Look at integer powers of $3$. We have $3^6=729$, so $3^7$ is about $2200$, bigger than $2013$.

For $e$, which is about $2.7$, we may need a bigger exponent, maybe $8$ or even $9$. We have that $3^8$ is about $6500$. And $(0.9)^8$ is therefore roughly $4\times 10^{-1}$. Multiply by $6500$. This puts us over $2013$. So exponent $8$ is too big, but closer than $7$.

Remark: In hindsight I should have worked directly with $2.7$. But the post describes how I actually calculated.

If you remember that $e^3 \approx 20$:

$$ln(2013) \approx ln(2000) = ln(20 \cdot 20 \cdot 5) = ln(20) + ln(20) + ln(5) \approx 3 + 3 + ln(5)$$

ln(5) is between 1 and 2 (because $e \approx 2.71$ and $e^2 \approx 7.4$), so all you need to known is if $ln(5)$ is greater or less than 1.5.

$e^{1.5} = \sqrt{e^3} \approx \sqrt{20} = \sqrt{4 \cdot 5} = 2 \cdot \sqrt{5} \approx 2 \cdot 2.2 = 4.4 < 5$, so $ln(5) > 1.5$.

=> $ln(2013) \approx 8$.